Let $n$ be a nonnegative integer. Prove that $\sum\limits_{r=1}^n \dfrac 1{r}\dbinom{n}{r} = \sum\limits_{r=1}^n \dfrac 1{r}(2^r - 1)$.
One thing I have tried is to represent both $\binom{n}{r}$ and $2^r$ as sums of binomial coefficients, i.e. $\sum \binom{i}{r-1}$ and $\sum \binom{r}{i}$ respectively, but it does not seem to be helpful. I have also tried to use binomial identities but I do not see how they can be applied to the problem.
$$\sum_{r=1}^n \frac{1}{r}\binom{n}{r}=\int_{0}^1\sum_{r=1}^n\binom{n}{r}x^{r-1}dx=\int_{0}^1\frac{(1+x)^r-1}{x}dx\\=\int_{0}^1\sum_{r=1}^n (1+x)^{r-1}dx\\=\sum_{r=1}^n \frac{2^r-1}{r}$$
Solution without using calculus:
$$\begin{align} \sum_{r=1}^n\frac 1r\binom nr &=\sum_{r=1}^n\frac 1r\sum_{k=r}^n\binom kr-\binom{k-1}r \qquad\qquad\qquad\text{(*)}\\ &=\sum_{r=1}^n\frac 1r\sum_{k=r}^n\binom {k-1}{r-1}\\ &=\sum_{r=1}^n\sum_{k=r}^n\frac1k\binom kr\\ &=\sum_{k=1}^n\frac 1k\sum_{r=1}^k\binom kr\\ &=\sum_{k=1}^n\frac 1k(2^k-1)\\ &=\sum_{r=1}^n\frac 1r(2^r-1)\qquad\blacksquare \end{align}$$ * by "untelescoping" $\binom nr$ and noting that $\binom {r-1}r=0$.
Just for seeing an elementary one:
Using Pascal's rule we have:
$$\sum _{k=1}^{n}\binom{n}{k}\frac{1}{k}=\sum _{k=1}^{n}\binom{n-1}{k-1}\frac{1}{k}+\underbrace{\sum _{k=1}^{n}\binom{n-1}{k}\frac{1}{k}}_{(1)}$$$$=\sum _{k=1}^{n}\binom{n-1}{k-1}\frac{1}{k}+\underbrace{\sum _{k=1}^{n}\binom{n-2}{k-1}\frac{1}{k}+\sum _{k=1}^{n}\binom{n-2}{k}\frac{1}{k}}_{(1)}$$$$=\sum _{k=1}^{n}\binom{n-1}{k-1}\frac{1}{k}+\sum _{k=1}^{n}\binom{n-2}{k-1}\frac{1}{k}+\sum _{k=1}^{n}\binom{n-3}{k-1}\frac{1}{k}+\sum _{k=1}^{n}\binom{n-3}{k}\frac{1}{k}$$ On the other hand: $$\sum _{k=1}^{n}\binom{n-r}{k-1}\frac{1}{k}=\frac{1}{n-r+1}\sum _{k=1}^{n}\binom{n-r+1}{k}$$$$=\frac{1}{n-r+1}\left[\color{red}{\sum _{k=0}^{n-r+1}\binom{n-r+1}{k}}+\sum _{k=n-r+2}^{n}\binom{n-r+1}{k}-1\right]$$$$=\frac{\color{red}{2^{n-r+1}}-1}{n-r+1}\tag{I}$$
Continuing this way:$$\sum _{k=1}^{n}\binom{n}{k}\frac{1}{k}=\sum _{k=1}^{n}\binom{n-1}{k-1}\frac{1}{k}+\sum _{k=1}^{n}\binom{n-2}{k-1}\frac{1}{k}+...+\sum _{k=1}^{n}\binom{n-(n-1)}{k-1}\frac{1}{k}+\color{blue}{\sum _{k=1}^{n}\binom{n-(n-1)}{k}\frac{1}{k}}$$
Using (I) implies:
$$=\sum_{k=0}^{n-2}\frac{2^{\left(n-k\right)}-1}{n-k}+\color{blue}{1}$$
Hence: $$\sum _{k=1}^{n}\binom{n}{k}\frac{1}{k}=\sum_{k=0}^{n-2}\frac{2^{\left(n-k\right)}-1}{n-k}+\color{blue}{1}$$
Setting $n-k \mapsto k$ follows:
$$\bbox[5px,border:2px solid #00A000]{\sum _{k=1}^{n}\binom{n}{k}\frac{1}{k}=\sum_{k=1}^{n}\frac{2^{k}-1}{k}}$$
Another solution written for one of my classes.
We must prove that \begin{equation} \sum_{r=1}^{n}\dfrac{1}{r}\dbinom{n}{r}=\sum_{r=1}^{n}\dfrac{2^{r}-1} {r} \label{darij1.eq.1} \tag{1} \end{equation} holds for each nonnegative integer $n$.
Here is a proof of \eqref{darij1.eq.1} by induction on $n$:
The base case (the case $n=0$) is trivial, since the equality \eqref{darij1.eq.1} boils down to $0=0$ in this case (recall that empty sums are $0$ by definition).
For the induction step, we fix some positive integer $m$, and we assume that \eqref{darij1.eq.1} holds for $n=m-1$. We must now prove that \eqref{darij1.eq.1} holds for $n=m$.
We recall the following basic facts:
Sum of a row in Pascal's triangle: We have \begin{equation} \sum_{r=0}^{n}\dbinom{n}{r}=2^{n}\qquad\text{for each integer }n\geq 0. \label{darij1.eq.2n} \tag{2} \end{equation} (This follows from substituting $x=1$ and $y=1$ into the binomial formula $\left( x+y\right) ^{n}=\sum\limits_{r=0}^{n}\dbinom{n}{r}x^{r}y^{n-r}$.)
Absorption identity: We have \begin{align} \dfrac{n}{k}\dbinom{n-1}{k-1}=\dbinom{n}{k}\qquad\text{for any integers }n\text{ and }k>0. \end{align} (This follows by recalling the definitions of the two binomial coefficients: \begin{align*} \dbinom{n}{k} & =\dfrac{n\left( n-1\right) \left( n-2\right) \cdots\left( n-k+1\right) }{k!}\qquad\text{and}\\ \dbinom{n-1}{k-1} & =\dfrac{\left( n-1\right) \left( n-2\right) \cdots\left( n-k+1\right) }{\left( k-1\right) !}, \end{align*} and comparing the left and right hand sides.)
Pascal recursion: We have \begin{align} \dbinom{n}{k}=\dbinom{n-1}{k-1}+\dbinom{n-1}{k}\qquad\text{for any integers }n\text{ and }k>0. \end{align}
Furthermore, recall that $\dbinom{n}{k}=0$ whenever $n$ is a nonnegative integer and $k$ is an integer satisfying $k>n$. Applying this to $n=m-1$ and $k=m$, we obtain $\dbinom{m-1}{m}=0$ (since $m-1$ is a nonnegative integer and $m>m-1$).
However, the Pascal recursion yields that \begin{align} \dbinom{m}{r}=\dbinom{m-1}{r-1}+\dbinom{m-1}{r} \end{align} for each integer $r>0$. Thus, \begin{align*} & \sum_{r=1}^{m}\dfrac{1}{r}\underbrace{\dbinom{m}{r}}_{=\dbinom{m-1} {r-1}+\dbinom{m-1}{r}}\\ & =\sum_{r=1}^{m}\dfrac{1}{r}\left( \dbinom{m-1}{r-1}+\dbinom{m-1}{r}\right) \\ & =\sum_{r=1}^{m}\dfrac{1}{m}\cdot\underbrace{\dfrac{m}{r}\dbinom{m-1}{r-1} }_{\substack{=\dbinom{m}{r}\\\text{(by the absorption identity)} }}+\underbrace{\sum_{r=1}^{m}\dfrac{1}{r}\dbinom{m-1}{r}}_{=\sum \limits_{r=1}^{m-1}\dfrac{1}{r}\dbinom{m-1}{r}+\dfrac{1}{m}\dbinom{m-1}{m}}\\ & =\underbrace{\sum_{r=1}^{m}\dfrac{1}{m}\cdot\dbinom{m}{r}}_{=\dfrac{1} {m}\sum\limits_{r=1}^{m}\dbinom{m}{r}}+\sum\limits_{r=1}^{m-1}\dfrac{1} {r}\dbinom{m-1}{r}+\dfrac{1}{m}\underbrace{\dbinom{m-1}{m}}_{\substack{=0}}\\ & =\dfrac{1}{m}\underbrace{\sum\limits_{r=1}^{m}\dbinom{m}{r}}_{=\sum \limits_{r=0}^{m}\dbinom{m}{r}-\dbinom{m}{0}}+\underbrace{\sum\limits_{r=1} ^{m-1}\dfrac{1}{r}\dbinom{m-1}{r}}_{\substack{=\sum\limits_{r=1}^{m-1} \dfrac{2^{r}-1}{r}\\\text{(because we assumed }\\\text{that \eqref{darij1.eq.1} holds for }n=m-1\text{)}}}+\underbrace{\dfrac{1}{m}0} _{=0}\\ & =\dfrac{1}{m}\left( \underbrace{\sum\limits_{r=0}^{m}\dbinom{m}{r} }_{\substack{=2^{m}\\\text{(by \eqref{darij1.eq.2n})}}}-\underbrace{\dbinom {m}{0}}_{=1}\right) +\sum\limits_{r=1}^{m-1}\dfrac{2^{r}-1}{r}\\ & =\underbrace{\dfrac{1}{m}\left( 2^{m}-1\right) }_{=\dfrac{2^{m}-1}{m} }+\sum\limits_{r=1}^{m-1}\dfrac{2^{r}-1}{r}=\dfrac{2^{m}-1}{m}+\sum \limits_{r=1}^{m-1}\dfrac{2^{r}-1}{r}\\ & =\sum\limits_{r=1}^{m}\dfrac{2^{r}-1}{r}. \end{align*} In other words, \eqref{darij1.eq.1} holds for $n=m$. This completes the induction step. Thus, \eqref{darij1.eq.1} is proved by induction.
Remark: Another known identity analogous to \eqref{darij1.eq.1} is \begin{equation} \sum_{r=1}^{n}\dfrac{\left(-1\right)^{r-1}}{r}\dbinom{n}{r} = \sum_{r=1}^{n}\dfrac{1}{r} . \label{darij1.eq.3} \tag{3} \end{equation} It is a nice exercise to unite \eqref{darij1.eq.1} and \eqref{darij1.eq.3} under a common generalization.